Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{x^3 + 9x^2 + 18x}{4x^3 - 4x^2 - 168x}$
First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {x(x^2 + 9x + 18)} {4x(x^2 - x - 42)} $ $ k = \dfrac{x}{4x} \cdot \dfrac{x^2 + 9x + 18}{x^2 - x - 42} $ Simplify: $ k = \dfrac{1}{4} \cdot \dfrac{x^2 + 9x + 18}{x^2 - x - 42}$ Since we are dividing by $x$ , we must remember that $x \neq 0$ Next factor the numerator and denominator. $ k = \dfrac{1}{4} \cdot \dfrac{(x + 6)(x + 3)}{(x + 6)(x - 7)}$ Assuming $x \neq -6$ , we can cancel the $x + 6$ $ k = \dfrac{1}{4} \cdot \dfrac{x + 3}{x - 7}$ Therefore: $ k = \dfrac{ x + 3 }{ 4(x - 7)}$, $x \neq -6$, $x \neq 0$